What are the constants a and b so that f=x^3+ax^2+bx+c have a relative maximum at x=-1 and relative min at x=3?
You need to remember that the critical values of a function express the zeroes of derivative of function, hence you need to find derivative of f(x).
`f'(x) = (x^3+ax^2+bx+c)' =gt f'(x) = 3x^2 + 2ax + b`
The problem provides you the following information: the function reaches a relative maximum at x = -1, hence f'(-1)=0
You need to substitute -1 for x...
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